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6.90 gm of a metal carbonate were dissolved in 60 ml of 2(N) HCl. The excess acid was neutralized by 20 ml of 1(N) NaOH. What is the equiv. alent wt. of metal ?

(1) 40 

(2) 20

(3) 19 

(4) 39

Sarthaks Test For Class 7-12

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answered ago by (20.1k points)
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Correct Answer is: (4) 39

Equiv. of HCl taken = 60 × 2 × 10–3

Equiv. of HCl present after the reaction = 20 × 1 × 10–3

∴ Equiv. of HCl utilized = 100 × 10–3

∴ 100 × 10–3 equiv. of metal carbonate = 6.90 gm

∴ 1 equiv. of metal carbonate

= 6.69/10-1 = 69 gm

∴ equiv. wt. of metal = 69 – 30 = 39

[becuase equiv. wt. of carbonate = 30]

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